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I've known, for $\beta > \alpha$, $\int_\alpha^\beta(x-\alpha)(x-\beta)dx=-\frac16(\beta-\alpha)^3$ for a while, and I've experimented with its "relatives" such as $$\int_\alpha^\beta(x-\alpha)^2(x-\beta)dx$$ $$\int_\alpha^\beta(x-\alpha)(x-\beta)^2dx$$ $$\int_\alpha^\beta(x-\alpha)^2(x-\beta)^2dx$$ and others. Then, I noticed a clear sequence. Each integrand is next to its respective integral from $\alpha$ to $\beta$.

$$(x-\alpha)^2(x-\beta) : -\frac{1}{12}(\beta-\alpha)^4$$ $$(x-\alpha)^2(x-\beta)^2 : \frac{1}{30}(\beta-\alpha)^5$$ $$(x-\alpha)^3(x-\beta)^2 : \frac{1}{60}(\beta-\alpha)^6$$ $$(x-\alpha)^3(x-\beta)^3 : -\frac{1}{140}(\beta-\alpha)^7$$ $$(x-\alpha)^4(x-\beta)^3 : -\frac{1}{280}(\beta-\alpha)^8$$ $$\vdots$$

I was concerned about the denominators of the coefficients of the integrals, $12, 30, 60, 140, 280, 630, 1260, 2772 \dots$, particularly about how they seem to have no relation with each other.

I am wondering whether this sequence is special, whether you could make this sequence without the painstaking integration, and whether it has applications beyond integrals.

Travis Willse
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sreysus
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  • A100071, perhaps? – kipf Jun 28 '23 at 19:56
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    In case you reconsider to have the integration approach: $$\int_a^b (x-a)^n(x-b)^kdx\overset{\large \frac{a-x}{a-b}\to x}=(a-b)^{n+k+1}\int_0^1 x^n(1-x)^kdx=(a-b)^{n+k+1}\frac{n! k!}{(n+k+1)!}$$ – Zacky Jun 28 '23 at 20:07
  • @Zacky beat me to it. OP, what you have is a generalization of what is known as the beta function (the case when $a = 0$ and $b = 1$. This is a very important function to statisticians. – Charles Hudgins Jun 30 '23 at 05:09

2 Answers2

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Applying the affine substitution $u = \frac{x - \alpha}{\beta - \alpha}$, $dx = \frac{du}{\beta - \alpha}$, transforms the integral to $$(-1)^k (\beta - \alpha)^{j + k + 1} \int_0^1 u^j (1 - u)^k \,du$$ so, there's no loss of generality if we consider $\alpha = 0, \beta = 1$ and consider just the integral, which also eliminates the pesky minus sign. Now, $$\int_0^1 u^j (1 - u)^k \,du = \mathrm{B}(j + 1, k + 1) = \frac{\Gamma(j + 1) \Gamma(k + 1)}{\Gamma(j + k + 2)} ,$$ where $\mathrm{B}$ and $\Gamma$ are the beta and gamma functions, respectively. If $j, k \in \Bbb N$, this expression specializes to $$\frac{j! k!}{(j + k + 1)!} = \frac{1}{(j + k + 1)} \frac{1}{{j + k} \choose j} .$$

Now, the sequence you've observed might be simpler to regard as one sequence formed by interweaving two different sequences, one term at a time.

The second is the sequence of values with $k = j$, namely, $$b_j := \int_0^1 u^j (1 - u)^j \,du = \frac{1}{(2 j + 1)!} \frac{1}{{2 j} \choose j} ,$$ and starting from $j = 0$ the reciprocals $\frac{1}{b_j}$ of the entries are $$ 1, 6, 30, 140, 630, 2772, 12012, 51480, 218790, 923780 .$$ This latter sequence is A002457 and appears in numerous other contexts, mostly combinatorial. Up to a natural choice of scaling factor this sequence gives the ratio of volumes of the $(2 j + 1)$ sphere and the $(2 j)$-sphere. See Peter Luschny's Oct 12 2015 comment in the link for details.

The first is the sequence of values with $k = j - 1$, namely $$c_j = \frac{j! (j - 1)!}{(2 j)!} = \frac{1}{(2j)} \frac{1}{{2j - 1} \choose j} ,$$ and we can see that $b_{j - 1} = 2 c_j$. The reciprocals $\frac{1}{c_j}$ are the Apéry numbers A005430: $$2, 12, 60, 280, 1260, 5544, 24024, 102960, 437580, 1847560, \ldots .$$ Asymptotically, we have $$c_j = \frac{\pi}{2^{2 j + 1} \sqrt{j}} + R_j,$$ where $R_j \in O(2^{-(2 j + 1)} j^{-3 / 2})$.

Interleaving $\{b_j\}$ and $\{c_j\}$ produces your sequence, the reciprocals of whose terms are $$1, 2, 6, 12, 30, 60, 140, 280, 630, 1260, \ldots$$ by setting $$d_m := \frac{1}{m} \frac{1}{{m - 1} \choose {\left\lfloor (m - 1) / 2 \right\rfloor}} .$$ As kipf pointed out, the sequence appears in the OEIS as A100071. Some combinatorial interpretations are given at the link.

Travis Willse
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Reduction Formula $$ \begin{aligned} I(j, k) & =\int_\alpha^\beta(x-\alpha)^j(x-\beta)^k d x \\ & =\frac{1}{j+1} \int_\alpha^\beta(x-\beta)^k d(x-\alpha)^{j+1} \\ & =\frac{1}{j+1}\left[(x-\alpha)^j(x-\beta)^{k+1}\right]_\alpha^\beta-\frac{k}{j+1} \int_\alpha^\beta(x-\alpha)^{j+1}(x-\beta)^{k-1}dx \\ & =-\frac{k}{j+1} I(j+1, k-1) \end{aligned} $$

Repeatedly using the formula, we have $$ \begin{aligned} I (j,k)& =(-1)^k \frac{k}{j+1} \cdot \frac{k-1}{j+2} \cdot \cdots \frac{k-(k-1)}{j+k} \cdot I(j+k, 0) \\ & =\frac{(-1)^k k ! j!}{(j+k) !} \cdot \int_\alpha^\beta(x-\alpha)^{j+k} d x \\ & =\frac{(-1)^k}{\left(\begin{array}{c} j+k \\ k \end{array}\right)}\left[\frac{(x-\alpha)^{k+j+1}}{j+k+1}\right]_\alpha^\beta \\ & =\frac{(-1)^k}{\left(\begin{array}{c} j+k \\ k \end{array}\right)} \frac{(\beta-\alpha)^{j+k+1}}{j+k+1} \end{aligned} $$ by which we know that the denominator of coefficient asked by OP is $$ \boxed{\left(\begin{array}{c} j+k \\ k \end{array}\right) (j+k+1)}$$ For examples, $$ \begin{aligned} & I(2,1)=\frac{(-1)^1(\beta-\alpha)^4}{\left(\begin{array}{l} 3 \\ 1 \end{array}\right)(4)}=-\frac{1}{12}(\beta-\alpha)^4 \\ & I(3,2)=\frac{(-1)^2(\beta-\alpha)^6}{\left(\begin{array}{l} 5 \\ 2 \end{array}\right)(6)}=\frac{1}{60}(\beta-\alpha)^6 \end{aligned} $$

Lai
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